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

1 Answer

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

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