Calculate the temperature of 4.0 mol of gas occupying $$5dm^3$$ at 3.32 bar. $$(R = 0.083\; bar\; dm^3 K^{–1} mol–^{-1})$$.

Asked by Pragya Singh | 1 year ago |  75

##### Solution :-

N= 4.0 mol

V = $$5 dm^{ 3 }$$

p = 3.32 bar

R = $$0.083\; bar\; dm^{ 3 }K^{-1} mol^{-1}$$

The temp (T) can be calculated using the ideal gas equation as:

pV = nRT

T =$$\dfrac{ pV }{ nR }$$

$$\dfrac{3.32 \; \times \; 5 }{ 4 \;\times \; 0.083 }$$

= 50 K

Therefore, the required temp is 50 K.

Answered by Abhisek | 1 year ago

### Related Questions

#### Explain the physical significance of Van der Waals parameters.

Explain the physical significance of Van der Waals parameters.

#### Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively.

Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?

#### In terms of Charles’ law explain why -273°C is the lowest possible temperature.

In terms of Charles’ law explain why -273°C is the lowest possible temperature.

What would be the SI units for the quantity $$\dfrac{pV^2 T^2}{n}$$